# Why do diode rings multiply?

Why do diode rings multiply signals?

I have been playing around with mixers. Home brew diode rings of 1N4148 and the NE602 chip (double-balanced mixer and oscillator).

I think I understand the basic maths of signal multiplication, and getting the sum and difference frequencies, but can someone explain to be why it is that a diode ring or a Gilbert cell, multiply the two inputs? Why wouldn't they just add? What's actually going on in the diodes that causes the output to be the product of the two inputs at any time t?

The electronics articles I have read seem to focus on the mechanics of the diode ring and how the left or right hand pair of diodes act as a switch, but I still don't understand why the two inputs should multiply.

I have missed something I'm sure.

• Start with the two cases where Lo=fo, RF=DC, IF=fo, alternatively Lo=fo, RF=fo, IF=DC Commented May 28, 2017 at 10:24
• This old joke explains it perfectly: gist.github.com/skanev/cc965abcdbe3cb7b43ca Commented May 28, 2017 at 18:04

## 4 Answers

can someone explain to be why it is that a diode ring or a Gilbert cell, multiply the 2 inputs?

I'm going to focus on the gilbert cell. Take a look at the characteristic of BJTs, JFETs and MOSFETs: -

And, in particular, concentrate on the parts of the characteristics where the voltage across the device (Vce or Vds) is low. Using the MOSFET for this example you can see that below 2 volts between drain and source, the V/I characteristic is like a variable resistor whose ratio of V to I (aka resistance) is controlled by the gate voltage.

So you have the basis of signal multiplication and that is what a gilbert cell utilizes. Sure, there are several transistors that make up a gilbert cell but this is to provide multiplication in all four quadrants that is largely unaffected by temperature variations.

It's a similar story when using diodes as multipliers. Consider the forward characteristic of a diode: -

Look at the slope of the characteristic - as voltage increases, for a small additional change in voltage there is a much bigger change in current due to that small voltage change. This is called the slope resistance of the diode. So, if you inject a small AC current onto the diode and raise or lower the DC conditions, that small AC current results in a lower or greater AC signal voltage being present - this is modulation at its most basic and modulation is multiplication.

One diode doesn't make a very good multiplier because the output signal also contains a DC offset (modulating voltage) and there are non-linearities in the resulting AC output signal. Having a second diode circuit that is only fed the same dc conditions, allows a subtractor circuit to remove the DC conditions leaving just the distorted AC signal. Replicating the dual diode signal but feeding it the inverse of the AC signal allows you to make a full four quadrant multiplier similar to a diode ring (ring modulator) and the distortions are much reduced.

The essence of a diode mixers is the non linear behaviour of a diode. If you build a mixer with resistors, instead of diodes, it would not "mix" (obviously !) as resistors are linear devices. With resistors the shape of the input signal will match the shape of the output signal except for some scaling (attenuation).

But when you build a proper mixer using diodes you do get signal distortion because the diodes are non-linear devices. Their behaviour (Current/Voltage relation) is sort of exponential (diode equation) so far from linear. Even if you'd keep the signals small they would still distort because of non-linearities.

The math behind this is (for me) easily explained by the Taylor expansion. Any function (including the transfer of a mixer) can be represented as a Taylor expansion series. Note how this expansion contains $x$, $x^2$, $x^3$ etc... Now this $x^2$ is the second harmonic of $x$. So If you'd input $x$ into a non-linear device, you'd get (depending on how it distorts the signal) $x^2$, $x^3$ etc at the output.

In a mixer you input two signals, RF and LO. You could do a Taylor expansion on that simply assuming that the diodes only add 2nd order distortion (to keep it simple). Then you would find that if you input $x$ and $y$ to that mixer, at the output you'd get $xy$ and $yx$. And these are the sum and difference frequencies !

• +1, but the simplest way to understand this is probably $$\mathrm{ln}(a·b) = \mathrm{ln}(a) + \mathrm{ln}(b)$$ So given you have an exponential function somewhere, adding leads to a product in result. Commented May 28, 2017 at 11:33
• A potential conflict-of-terms: Audio folks desire a very linear mixer, where two signals combine with no by-products - a very different animal. RF folks take mixing as multiplying, where the desired outputs are the ones that audio folks absolutely don't want. Commented May 28, 2017 at 17:35
• @glen_geek I agree with you. It is actually "summing" what the audio folks want to do, not mixing. But everyone in that world calls it mixing. Also, mixing is really multiplying so the RF people (myself included) are actually also using the wrong term ;-) Commented May 28, 2017 at 17:43

The most simple explanation of a diode ring mixer is when you assume a large rectangular LO signal:

• during first LO half cycle one diode pair is forward biased; the other pair is reverse biased. I.e. you can treat one pair as a short and the other as an open circuit.
• during second half LO cycle it's vice versa.

Depending on which diode pair is conducting the input signal is transfered to the output either inverted or non-inverted.

In other words: the mixer creates an output signal that is the input signal multiplied by a +1/-1 rectangular LO signal.

If you look at the frequency spectrum of such an output signal (using Laplace or Fourier theory) you find that the most significant (first order) terms are exactly what you expect from a mixer:

$f_{out} = f_{in} \pm f_{LO}$.

(smaller, higher order terms get filtered out; as you need some kind of filter after the mixer anyway)

With signals strong enough to fully turn the diodes on and off (0.2 volts or more), then when one diode is off, the node gets switched to another signal, perhaps the 180-degree phase (or in polyphase mixers, some 360/N phase choice, where useful cancellation occurs).

• What if I told you that you don't need a strong enough signal to turn the diodes on and off yet the mixer will stil "mix". The non-linearity is also present when the diodes are DC-biased at a certain value and the small AC signals on top of that bias will cause be mixed due to non-linearities. The conversion gain might be lower though compared to driving the diodes "hard" but still, mixing will occur. Commented May 28, 2017 at 17:41
• Certainly. Used in that famous Varactor Input Bridge Opamp, that paid for the other experiments of the first employer of Bob Pease: Philbrick Nexus. Commented May 28, 2017 at 22:12
• @FakeMoustache You are correct, mixing will occur with just a single diode but that is not why the 4 diode ring is selected. I do believe that one of the signals (the LO usually) in a diode ring mixer is intended to be Vf*2 greater than the peak of the other signal of interest to achieve the benefits of the crude unamplified switching from the diodes. Commented May 29, 2017 at 6:03